Abstract

We introduce an iterative for finding the zeros point of the sum of two monotone operators. We prove that the suggested method converges strongly to the zeros point of the sum of two monotone operators.

1. Introduction

Let 𝐶 be a nonempty closed convex subset of a real Hilbert space 𝐻. Let 𝐴∶𝐶→𝐻 be a single-valued nonlinear mapping and let 𝐵∶𝐻→2𝐻 be a multivalued mapping. The “so-called” quasi-variational inclusion problem is to find a 𝑢∈2𝐻 such that0∈𝐴𝑥+𝐵𝑥.(1.1)
The set of solutions of (1.1) is denoted by (𝐴+𝐵)−1(0). A number of problems arising in structural analysis, mechanics, and economics can be studied in the framework of this kind of variational inclusions; see, for instance, [1–4]. The problem (1.1) includes many problems as special cases.(1)If 𝐵=𝜕𝜙∶𝐻→2𝐻, where 𝜙∶𝐻→𝑅∪+∞ is a proper convex lower semicontinuous function and 𝜕𝜙 is the subdif and if onlyerential of 𝜙, then the variational inclusion problem (1.1) is equivalent to find 𝑢∈𝐻 such that
⟨𝐴𝑢,𝑦−𝑢⟩+𝜙(𝑦)−𝜙(𝑢)≥0,∀𝑦∈𝐻,(1.2)
which is called the mixed quasi-variational inequality (see, Noor [5]).(2)If 𝐵=𝜕𝛿𝐶, where 𝐶 is a nonempty closed convex subset of 𝐻 and 𝛿𝐶∶𝐻→[0,∞] is the indicator function of 𝐶, that is,
𝛿𝐶=0,𝑥∈𝐶,+∞,𝑥∉𝐶,(1.3)
then the variational inclusion problem (1.1) is equivalent to find 𝑢∈𝐶 such that
⟨𝐴𝑢,𝑣−𝑢⟩≥0,∀𝑣∈𝐶.(1.4)

This problem is called Hartman-Stampacchia variational inequality (see, e.g., [6]).

Recently, Zhang et al. [7] introduced a new iterative scheme for finding a common element of the set of solutions to the inclusion problem, and the set of fixed points of nonexpansive mappings in Hilbert spaces. Peng et al. [8] introduced another iterative scheme by the viscosity approximate method for finding a common element of the set of solutions of a variational inclusion with set-valued maximal monotone mapping and inverse strongly monotone mappings, the set of solutions of an equilibrium problem, and the set of fixed points of a nonexpansive mapping. For some related works, please see [9–27] and the references therein.

Inspired and motivated by the works in the literature, in this paper, we introduce an iterative for solving the problem (1.1). We prove that the suggested method converges strongly to the zeros point of the sum of two monotone operators 𝐴+𝐵.

2. Preliminaries

Let 𝐻 be a real Hilbert space with inner product ⟨⋅,⋅⟩ and norm ‖⋅‖, respectively. Let 𝐶 be a nonempty closed convex subset of 𝐻. Recall that a mapping 𝐴∶𝐶→𝐻 is said to be 𝛼-inverse strongly-monotone if and if only ⟨𝐴𝑥−𝐴𝑦,𝑥−𝑦⟩≥𝛼‖𝐴𝑥−𝐴𝑦‖2(2.1)
for some 𝛼>0 and for all 𝑥,𝑦∈𝐶. It is known that if 𝐴 is 𝛼-inverse strongly monotone, then 1‖𝐴𝑥−𝐴𝑦‖≤𝛼‖𝑥−𝑦‖(2.2)
for all 𝑥,𝑦∈𝐶.

Let 𝐵 be a mapping of 𝐻 into 2𝐻. The effective domain of 𝐵 is denoted by dom(𝐵), that is,dom(𝐵)={𝑥∈𝐻∶𝐵𝑥≠∅}.(2.3)
A multivalued mapping 𝐵 is said to be a monotone operator on 𝐻 if and if only ⟨𝑥−𝑦,𝑢−𝑣⟩≥0(2.4)
for all 𝑥,𝑦∈dom(𝐵), 𝑢∈𝐵𝑥, and 𝑣∈𝐵𝑦. A monotone operator 𝐵 on 𝐻 is said to be maximal if and if only its graph is not strictly contained in the graph of any other monotone operator on 𝐻. Let 𝐵 be a maximal monotone operator on 𝐻 and let 𝐵−10={𝑥∈𝐻∶0∈𝐵𝑥}.

For a maximal monotone operator 𝐵 on 𝐻 and 𝜆>0, we may define a single-valued operator: 𝐽𝐵𝜆=(𝐼+𝜆𝐵)−1∶𝐻→dom(𝐵),(2.5)
which is called the resolvent of 𝐵 for 𝜆. It is known that the resolvent 𝐽𝐵𝜆 is firmly nonexpansive, that is, ‖‖𝐽𝐵𝜆𝑥−𝐽𝐵𝜆𝑦‖‖2≤𝐽𝐵𝜆𝑥−𝐽𝐵𝜆𝑦,𝑥−𝑦(2.6)
for all 𝑥,𝑦∈𝐶 and 𝐵−10=𝐹(𝐽𝐵𝜆) for all 𝜆>0.

The following resolvent identity is well known: for 𝜆>0 and 𝜇>0, there holds the following identity:𝐽𝐵𝜆𝑥=𝐽𝐵𝜇𝜇𝜆𝜇𝑥+1−𝜆𝐽𝐵𝜆𝑥,𝑥∈𝐻.(2.7)

We use the following notation: (i)𝑥𝑛⇀𝑥 stands for the weak convergence of (𝑥𝑛) to 𝑥; (ii)𝑥𝑛→𝑥 stands for the strong convergence of (𝑥𝑛) to 𝑥.

We need the following lemmas for the next section.

Lemma 2.1 (see [28]). Let 𝐶 be a nonempty closed convex subset of a real Hilbert space 𝐻. Let the mapping 𝐴∶𝐶→𝐻 be 𝛼-inverse strongly monotone and let 𝜆>0 be a constant. Then, one has
‖‖(𝐼−𝜆𝐴)𝑥−(𝐼−𝜆𝐴)𝑦2≤‖𝑥−𝑦‖2+𝜆(𝜆−2𝛼)‖𝐴𝑥−𝐴𝑦‖2,∀𝑥,𝑦∈𝐶.(2.8)
In particular, if 0≤𝜆≤2𝛼, then 𝐼−𝜆𝐴 is nonexpansive.

Lemma 2.2 (see [29]). Let {𝑥𝑛} and {𝑦𝑛} be bounded sequences in a Banach space 𝑋 and let {𝛽𝑛} be a sequence in [0,1] with
0<liminf𝑛→∞𝛽𝑛≤limsup𝑛→∞𝛽𝑛<1.(2.9)
Suppose that
𝑥𝑛+1=1−𝛽𝑛𝑦𝑛+𝛽𝑛𝑥𝑛(2.10)
for all 𝑛≥0 and
limsup𝑛→∞‖‖𝑦𝑛+1−𝑦𝑛‖‖−‖‖𝑥𝑛+1−𝑥𝑛‖‖≤0.(2.11)
Then, lim𝑛→∞‖𝑦𝑛−𝑥𝑛‖=0.

Lemma 2.3 (see [30]). Assume that {𝑎𝑛} is a sequence of nonnegative real numbers such that
𝑎𝑛+1≤1−𝛾𝑛𝑎𝑛+𝛿𝑛𝛾𝑛,(2.12)
where {𝛾𝑛} is a sequence in (0,1) and {𝛿𝑛} is a sequence such that (1)∑∞𝑛=1𝛾𝑛=∞; (2)limsup𝑛→∞𝛿𝑛≤0 or ∑∞𝑛=1|𝛿𝑛𝛾𝑛|<∞. Then lim𝑛→∞𝑎𝑛=0.

3. Main Results

In this section, we will prove our main result.

Theorem 3.1. Let 𝐶 be a nonempty closed and convex subset of a real Hilbert space 𝐻. Let 𝐴 be an 𝛼-inverse strongly monotone mapping of 𝐶 into H and let 𝐵 be a maximal monotone operator on 𝐻, such that the domain of 𝐵 is included in 𝐶. Let 𝐽𝐵𝜆=(𝐼+𝜆𝐵)−1 be the resolvent of 𝐵 for 𝜆>0. Suppose that (𝐴+𝐵)−10≠∅. For 𝑢∈𝐶 and given 𝑥0∈𝐶, let {𝑥𝑛}⊂𝐶 be a sequence generated by
𝑥𝑛+1=𝛽𝑛𝑥𝑛+1−𝛽𝑛𝐽𝐵𝜆𝑛𝛼𝑛𝑢+1−𝛼𝑛𝑥𝑛−𝜆𝑛𝐴𝑥𝑛(3.1)
for all 𝑛≥0, where {𝜆𝑛}⊂(0,2𝛼), {𝛼𝑛}⊂(0,1), and {𝛽𝑛}⊂(0,1) satisfy (i)lim𝑛→∞𝛼𝑛=0 and ∑𝑛𝛼𝑛=∞; (ii)0<liminf𝑛→∞𝛽𝑛≤limsup𝑛→∞𝛽𝑛<1; (iii)𝑎≤𝜆𝑛≤𝑏 where [𝑎,𝑏]⊂(0,2𝛼) and lim𝑛→∞(𝜆𝑛+1−𝜆𝑛)=0. Then {𝑥𝑛} generated by (3.1) converges strongly to ̃𝑥=𝑃(𝐴+𝐵)−10(𝑢).

Proof. First, we choose any 𝑧∈(𝐴+𝐵)−10. Note that
𝑧=𝐽𝐵𝜆𝑛𝑧−𝜆𝑛1−𝛼𝑛𝐴𝑧=𝐽𝐵𝜆𝑛𝛼𝑛𝑧+1−𝛼𝑛𝑧−𝜆𝑛𝐴𝑧(3.2)
for all 𝑛≥0. Since 𝐽𝐵𝜆 is nonexpansive for all 𝜆>0, we have
‖‖𝐽𝐵𝜆𝑛𝛼𝑛𝑢+1−𝛼𝑛𝑥𝑛−𝜆𝑛𝐴𝑥𝑛‖‖−𝑧2=‖‖𝐽𝐵𝜆𝑛𝛼𝑛𝑢+1−𝛼𝑛𝑥𝑛−𝜆𝑛𝐴𝑥𝑛−𝐽𝐵𝜆𝑛𝛼𝑛𝑧+1−𝛼𝑛𝑧−𝜆𝑛‖‖𝐴𝑧2≤‖‖𝛼𝑛𝑢+1−𝛼𝑛𝑥𝑛−𝜆𝑛𝐴𝑥𝑛−𝛼𝑛𝑧+1−𝛼𝑛𝑧−𝜆𝑛‖‖𝐴𝑧2=‖‖1−𝛼𝑛𝑥𝑛−𝜆𝑛𝐴𝑥𝑛−𝑧−𝜆𝑛𝐴𝑧+𝛼𝑛‖‖(𝑢−𝑧)2.(3.3)
Since 𝐴 is 𝛼-inverse strongly monotone, we get
‖‖1−𝛼𝑛𝑥𝑛−𝜆𝑛𝐴𝑥𝑛−𝑧−𝜆𝑛𝐴𝑧+𝛼𝑛‖‖(𝑢−𝑧)2≤1−𝛼𝑛‖‖𝑥𝑛−𝜆𝑛𝐴𝑥𝑛−𝑧−𝜆𝑛‖‖𝐴𝑧2+𝛼𝑛‖𝑢−𝑧‖2=1−𝛼𝑛‖‖𝑥𝑛−𝑧−𝜆𝑛𝐴𝑥𝑛‖‖−𝐴𝑧2+𝛼𝑛‖𝑢−𝑧‖2=1−𝛼𝑛‖‖𝑥𝑛‖‖−𝑧2−2𝜆𝑛⟨𝐴𝑥𝑛−𝐴𝑧,𝑥𝑛−𝑧⟩+𝜆2𝑛‖‖𝐴𝑥𝑛‖‖−𝐴𝑧2+𝛼𝑛‖𝑢−𝑧‖2≤1−𝛼𝑛‖‖𝑥𝑛‖‖−𝑧2−2𝛼𝜆𝑛‖‖𝐴𝑥𝑛‖‖−𝐴𝑧2+𝜆2𝑛‖‖𝐴𝑥𝑛‖‖−𝐴𝑧2+𝛼𝑛‖𝑢−𝑧‖2=1−𝛼𝑛‖‖𝑥𝑛‖‖−𝑧2+𝜆𝑛𝜆𝑛‖‖−2𝛼𝐴𝑥𝑛‖‖−𝐴𝑧2+𝛼𝑛‖𝑢−𝑧‖2.(3.4)
By (3.3) and (3.4), we obtain
‖‖𝐽𝐵𝜆𝑛𝛼𝑛𝑢+1−𝛼𝑛𝑥𝑛−𝜆𝑛𝐴𝑥𝑛‖‖−𝑧2≤1−𝛼𝑛‖‖𝑥𝑛‖‖−𝑧2+𝜆𝑛𝜆𝑛‖‖−2𝛼𝐴𝑥𝑛‖‖−𝐴𝑧2+𝛼𝑛‖𝑢−𝑧‖2≤1−𝛼𝑛‖‖𝑥𝑛‖‖−𝑧2+𝛼𝑛‖𝑢−𝑧‖2.(3.5)
It follows from (3.1) and (3.5) that
‖‖𝑥𝑛+1‖‖−𝑧2=‖‖𝛽𝑛𝑥𝑛+−𝑧1−𝛽𝑛𝐽𝐵𝜆𝑛𝛼𝑛𝑢+1−𝛼𝑛𝑥𝑛−𝜆𝑛𝐴𝑥𝑛‖‖−𝑧2≤𝛽𝑛‖‖𝑥𝑛‖‖−𝑧2+1−𝛽𝑛‖‖𝐽𝐵𝜆𝑛𝛼𝑛𝑢+1−𝛼𝑛𝑥𝑛−𝜆𝑛𝐴𝑥𝑛‖‖−𝑧2≤𝛽𝑛‖‖𝑥𝑛‖‖−𝑧2+1−𝛽𝑛1−𝛼𝑛‖‖𝑥𝑛‖‖−𝑧2+𝛼𝑛‖𝑢−𝑧‖2=1−1−𝛽𝑛𝛼𝑛‖‖𝑥𝑛‖‖−𝑧2+1−𝛽𝑛𝛼𝑛‖𝑢−𝑧‖2‖‖𝑥≤max𝑛‖‖−𝑧2,‖𝑢−𝑧‖2.(3.6)
By induction, we have
‖‖𝑥𝑛+1‖‖‖‖𝑥−𝑧≤max0‖‖−𝑧,‖𝑢−𝑧‖.(3.7)
Therefore, {𝑥𝑛} is bounded. We deduce immediately that {𝐴𝑥𝑛} is also bounded. Set 𝑢𝑛=𝛼𝑛𝑢+(1−𝛼𝑛)(𝑥𝑛−𝜆𝑛𝐴𝑥𝑛) for all 𝑛. Then {𝑢𝑛} and {𝐽𝐵𝜆𝑛𝑢𝑛} are bounded.Next, we estimate ‖𝐽𝐵𝜆𝑛+1𝑢𝑛+1−𝐽𝐵𝜆𝑛𝑢𝑛‖. In fact, we have
‖‖𝐽𝐵𝜆𝑛+1𝑢𝑛+1−𝐽𝐵𝜆𝑛𝑢𝑛‖‖=‖‖𝐽𝐵𝜆𝑛+1𝛼𝑛+1𝑢+1−𝛼𝑛+1𝑥𝑛+1−𝜆𝑛+1𝐴𝑥𝑛+1−𝐽𝐵𝜆𝑛𝛼𝑛𝑢+1−𝛼𝑛𝑥𝑛−𝜆𝑛𝐴𝑥𝑛‖‖≤‖‖𝐽𝐵𝜆𝑛+1𝛼𝑛+1𝑢+1−𝛼𝑛+1𝑥𝑛+1−𝜆𝑛+1𝐴𝑥𝑛+1−𝐽𝐵𝜆𝑛+1𝛼𝑛𝑢+1−𝛼𝑛𝑥𝑛−𝜆𝑛𝐴𝑥𝑛‖‖+‖‖𝐽𝐵𝜆𝑛+1𝛼𝑛𝑢+1−𝛼𝑛𝑥𝑛−𝜆𝑛𝐴𝑥𝑛−𝐽𝐵𝜆𝑛𝛼𝑛𝑢+1−𝛼𝑛𝑥𝑛−𝜆𝑛𝐴𝑥𝑛‖‖≤‖‖𝛼𝑛+1𝑢+1−𝛼𝑛+1𝑥𝑛+1−𝜆𝑛+1𝐴𝑥𝑛+1−𝛼𝑛𝑢+1−𝛼𝑛𝑥𝑛−𝜆𝑛𝐴𝑥𝑛‖‖+‖‖𝐽𝐵𝜆𝑛+1𝑢𝑛−𝐽𝐵𝜆𝑛𝑢𝑛‖‖≤‖‖𝐼−𝜆𝑛+1𝐴𝑥𝑛+1−𝐼−𝜆𝑛+1𝐴𝑥𝑛‖‖+||𝜆𝑛+1−𝜆𝑛||‖‖𝐴𝑥𝑛‖‖+𝛼𝑛+1‖‖𝑥‖𝑢‖+𝑛+1‖‖+𝜆𝑛+1‖‖𝐴𝑥𝑛+1‖‖+𝛼𝑛‖‖𝑥‖𝑢‖+𝑛‖‖+𝜆𝑛‖‖𝐴𝑥𝑛‖‖+‖‖𝐽𝐵𝜆𝑛+1𝑢𝑛−𝐽𝐵𝜆𝑛𝑢𝑛‖‖.(3.8)
Since 𝐼−𝜆𝑛+1𝐴 is nonexpansive for 𝜆𝑛+1∈(0,2𝛼), we have ‖(𝐼−𝜆𝑛+1𝐴)𝑥𝑛+1−(𝐼−𝜆𝑛+1𝐴)𝑥𝑛‖≤‖𝑥𝑛+1−𝑥𝑛‖. By the resolvent identity (2.7), we have
𝐽𝐵𝜆𝑛+1𝑢𝑛=𝐽𝐵𝜆𝑛𝜆𝑛𝜆𝑛+1𝑢𝑛+𝜆1−𝑛𝜆𝑛+1𝐽𝐵𝜆𝑛+1𝑢𝑛.(3.9)
It follows that
‖‖𝐽𝐵𝜆𝑛+1𝑢𝑛−𝐽𝐵𝜆𝑛𝑢𝑛‖‖=‖‖‖𝐽𝐵𝜆𝑛𝜆𝑛𝜆𝑛+1𝑢𝑛+𝜆1−𝑛𝜆𝑛+1𝐽𝐵𝜆𝑛+1𝑢𝑛−𝐽𝐵𝜆𝑛𝑢𝑛‖‖‖≤‖‖‖𝜆𝑛𝜆𝑛+1𝑢𝑛+𝜆1−𝑛𝜆𝑛+1𝐽𝐵𝜆𝑛+1𝑢𝑛−𝑢𝑛‖‖‖≤||𝜆𝑛+1−𝜆𝑛||𝜆𝑛+1‖‖𝑢𝑛−𝐽𝐵𝜆𝑛+1𝑢𝑛‖‖.(3.10)
So,
‖‖𝐽𝐵𝜆𝑛+1𝑢𝑛+1−𝐽𝐵𝜆𝑛𝑢𝑛‖‖≤‖‖𝑥𝑛+1−𝑥𝑛‖‖+||𝜆𝑛+1−𝜆𝑛||‖‖𝐴𝑥𝑛‖‖+𝛼𝑛+1‖‖𝑥‖𝑢‖+𝑛+1‖‖+𝜆𝑛+1‖‖𝐴𝑥𝑛+1‖‖+𝛼𝑛‖‖𝑥‖𝑢‖+𝑛‖‖+𝜆𝑛‖‖𝐴𝑥𝑛‖‖+||𝜆𝑛+1−𝜆𝑛||𝜆𝑛+1‖‖𝑢𝑛−𝐽𝐵𝜆𝑛+1𝑢𝑛‖‖.(3.11)
Thus,
limsup𝑛→∞‖‖𝐽𝐵𝜆𝑛+1𝑢𝑛+1−𝐽𝐵𝜆𝑛𝑢𝑛‖‖−‖‖𝑥𝑛+1−𝑥𝑛‖‖≤0.(3.12)
From Lemma 2.2, we get
lim𝑛→∞‖‖𝐽𝐵𝜆𝑛𝑢𝑛−𝑥𝑛‖‖=0.(3.13)
Consequently, we obtain
lim𝑛→∞‖‖𝑥𝑛+1−𝑥𝑛‖‖=lim𝑛→∞1−𝛽𝑛‖‖𝐽𝐵𝜆𝑛𝑢𝑛−𝑥𝑛‖‖=0.(3.14)
From (3.5) and (3.6), we have
‖‖𝑥𝑛+1‖‖−𝑧2≤𝛽𝑛‖‖𝑥𝑛‖‖−𝑧2+1−𝛽𝑛‖‖𝐽𝐵𝜆𝑛𝛼𝑛𝑢+1−𝛼𝑛𝑥𝑛−𝜆𝑛𝐴𝑥𝑛‖‖−𝑧2≤1−𝛽𝑛1−𝛼𝑛‖‖𝑥𝑛‖‖−𝑧2+𝜆𝑛𝜆𝑛‖‖−2𝛼𝐴𝑥𝑛‖‖−𝐴𝑧2+𝛼𝑛‖𝑢−𝑧‖2+𝛽𝑛‖‖𝑥𝑛‖‖−𝑧2=1−1−𝛽𝑛𝛼𝑛‖‖𝑥𝑛‖‖−𝑧2+1−𝛽𝑛𝜆𝑛𝜆𝑛‖‖−2𝛼𝐴𝑥𝑛‖‖−𝐴𝑧2+1−𝛽𝑛𝛼𝑛‖𝑢−𝑧‖2≤‖‖𝑥𝑛‖‖−𝑧2+1−𝛽𝑛𝜆𝑛𝜆𝑛‖‖−2𝛼𝐴𝑥𝑛‖‖−𝐴𝑧2+1−𝛽𝑛𝛼𝑛‖𝑢−𝑧‖2.(3.15)
It follows that
1−𝛽𝑛𝜆𝑛2𝛼−𝜆𝑛‖‖𝐴𝑥𝑛‖‖−𝐴𝑧2≤‖‖𝑥𝑛‖‖−𝑧2−‖‖𝑥𝑛+1‖‖−𝑧2+1−𝛽𝑛𝛼𝑛‖𝑢−𝑧‖2≤‖‖𝑥𝑛‖‖−‖‖𝑥−𝑧𝑛+1‖‖‖‖𝑥−𝑧𝑛+1−𝑥𝑛‖‖+1−𝛽𝑛𝛼𝑛‖𝑢−𝑧‖2.(3.16)
Since lim𝑛→∞𝛼𝑛=0, lim𝑛→∞‖𝑥𝑛+1−𝑥𝑛‖=0, and liminf𝑛→∞(1−𝛽𝑛)𝜆𝑛(2𝛼−𝜆𝑛)>0, we have
lim𝑛→∞‖‖𝐴𝑥𝑛‖‖−𝐴𝑧=0.(3.17)
Put ̃𝑥=𝑃(𝐴+𝐵)−10(𝑢). Set 𝑣𝑛=𝑥𝑛−(𝜆𝑛/(1−𝛼𝑛))(𝐴𝑥𝑛−𝐴̃𝑥) for all 𝑛. Take 𝑧=̃𝑥 in (3.17) to get ‖𝐴𝑥𝑛−𝐴̃𝑥‖→0. First, we prove limsup𝑛→∞⟨𝑢−̃𝑥,𝑣𝑛−̃𝑥⟩≤0. We take a subsequence {𝑣𝑛𝑖} of {𝑣𝑛} such that
limsup𝑛→∞⟨𝑢−̃𝑥,𝑣𝑛−̃𝑥⟩=lim𝑖→∞𝑢−̃𝑥,𝑣𝑛𝑖.−̃𝑥(3.18)
It is clear that {𝑣𝑛𝑖} is bounded due to the boundedness of {𝑥𝑛} and ‖𝐴𝑥𝑛−𝐴̃𝑥‖→0. Then, there exists a subsequence {𝑣𝑛𝑖𝑗} of {𝑣𝑛𝑖} which converges weakly to some point 𝑤∈𝐶. Hence, {𝑥𝑛𝑖𝑗} also converges weakly to 𝑤 because of ‖𝑣𝑛𝑖𝑗−𝑥𝑛𝑖𝑗‖→0. By the similar argument as that in [31], we can show that 𝑤∈(𝐴+𝐵)−10. This implies that
limsup𝑛→∞⟨𝑢−̃𝑥,𝑣𝑛−̃𝑥⟩=lim𝑗→∞𝑢−̃𝑥,𝑣𝑛𝑖𝑗−̃𝑥=⟨𝑢−̃𝑥,𝑤−̃𝑥⟩.(3.19)
Note that ̃𝑥=𝑃(𝐴+𝐵)−10(𝑢). Then, ⟨𝑢−̃𝑥,𝑤−̃𝑥⟩≤0,𝑤∈(𝐴+𝐵)−10. Therefore,
limsup𝑛→∞⟨𝑢−̃𝑥,𝑣𝑛−̃𝑥⟩≤0.(3.20)
Finally, we prove that 𝑥𝑛→̃𝑥. From (3.1), we have
‖‖𝑥𝑛+1‖‖−̃𝑥2≤𝛽𝑛‖‖𝑥𝑛‖‖−̃𝑥2+1−𝛽𝑛‖‖𝐽𝐵𝜆𝑛𝑢𝑛‖‖−̃𝑥2=𝛽𝑛‖‖𝑥𝑛‖‖−̃𝑥2+1−𝛽𝑛‖‖𝐽𝐵𝜆𝑛𝑢𝑛−𝐽𝐵𝜆𝑛̃𝑥−1−𝛼𝑛𝜆𝑛‖‖𝐴̃𝑥2≤𝛽𝑛‖‖𝑥𝑛‖‖−̃𝑥2+1−𝛽𝑛‖‖𝑢𝑛−̃𝑥−1−𝛼𝑛𝜆𝑛‖‖𝐴̃𝑥2=𝛽𝑛‖‖𝑥𝑛‖‖−̃𝑥2+1−𝛽𝑛‖‖𝛼𝑛𝑢+1−𝛼𝑛𝑥𝑛−𝜆𝑛𝐴𝑥𝑛−̃𝑥−1−𝛼𝑛𝜆𝑛‖‖𝐴̃𝑥2=𝛽𝑛‖‖𝑥𝑛‖‖−̃𝑥2+1−𝛽𝑛‖‖1−𝛼𝑛𝑥𝑛−𝜆𝑛𝐴𝑥𝑛−̃𝑥−𝜆𝑛𝐴̃𝑥+𝛼𝑛‖‖(𝑢−̃𝑥)2=𝛽𝑛‖‖𝑥𝑛‖‖−̃𝑥2+1−𝛽𝑛×1−𝛼𝑛2‖‖𝑥𝑛−𝜆𝑛𝐴𝑥𝑛−̃𝑥−𝜆𝑛‖‖𝐴̃𝑥2+2𝛼𝑛1−𝛼𝑛𝑥𝑢−̃𝑥,𝑛−𝜆𝑛𝐴𝑥𝑛−̃𝑥−𝜆𝑛𝐴̃𝑥+𝛼2𝑛‖𝑢−̃𝑥‖2≤𝛽𝑛‖‖𝑥𝑛‖‖−̃𝑥2+1−𝛽𝑛×1−𝛼𝑛‖‖𝑥𝑛‖‖−̃𝑥2+2𝛼𝑛1−𝛼𝑛𝑢−̃𝑥,𝑥𝑛−𝜆𝑛𝐴𝑥𝑛−𝐴̃𝑥−̃𝑥+𝛼2𝑛‖𝑢−̃𝑥‖2≤1−1−𝛽𝑛𝛼𝑛‖‖𝑥𝑛‖‖−̃𝑥2+1−𝛽𝑛𝛼𝑛21−𝛼𝑛⟨𝑢−̃𝑥,𝑣𝑛−̃𝑥⟩+𝛼𝑛‖𝑢−̃𝑥‖2.(3.21)
It is clear that ∑𝑛(1−𝛽𝑛)𝛼𝑛=∞ and limsup𝑛→∞(2(1−𝛼𝑛)⟨𝑢−̃𝑥,𝑣𝑛−̃𝑥⟩+𝛼𝑛‖𝑢−̃𝑥‖2)≤0. We can therefore apply Lemma 2.3 to conclude that 𝑥𝑛→̃𝑥. This completes the proof.

4. Applications

Next, we consider the problem for finding the minimum norm solution of a mathematical model related to equilibrium problems. Let 𝐶 be a nonempty, closed, and convex subset of a Hilbert space and let 𝐺∶𝐶×𝐶→𝑅 be a bifunction satisfying the following conditions:(E1)𝐺(𝑥,𝑥)=0 for all 𝑥∈𝐶;(E2)𝐺 is monotone, that is, 𝐺(𝑥,𝑦)+𝐺(𝑦,𝑥)≤0 for all 𝑥,𝑦∈𝐶;(E3)for all 𝑥,𝑦,𝑧∈𝐶, limsup𝑡↓0𝐺(𝑡𝑧+(1−𝑡)𝑥,𝑦)≤𝐺(𝑥,𝑦);(E4)for all 𝑥∈𝐶, 𝐺(𝑥,⋅) is convex and lower semicontinuous.

Then, the mathematical model related to equilibrium problems (with respect to 𝐶) is to find ̃𝑥∈𝐶 such that𝐺(̃𝑥,𝑦)≥0(4.1)
for all 𝑦∈𝐶. The set of such solutions ̃𝑥 is denoted by 𝐸𝑃(𝐺). The following lemma appears implicitly in Blum and Oettli [32].

Lemma 4.1. Let 𝐶 be a nonempty, closed, and convex subset of 𝐻 and let 𝐺 be a bifunction of 𝐶×𝐶 into 𝑅 satisfying (E1)–(E4). Let 𝑟>0 and 𝑥∈𝐻. Then, there exists 𝑧∈𝐶 such that
1𝐺(𝑧,𝑦)+𝑟⟨𝑦−𝑧,𝑧−𝑥⟩≥0,∀𝑦∈C.(4.2)

Lemma 4.2. Assume that 𝐺∶𝐶×𝐶→𝑅 satisfies (E1)–(E4). For 𝑟>0 and 𝑥∈𝐻, define a mapping 𝑇𝑟∶𝐻→𝐶 as follows:
𝑇𝑟1(𝑥)=𝑧∈𝐶∶𝐺(𝑧,𝑦)+𝑟⟨𝑦−𝑧,𝑧−𝑥⟩≥0,∀𝑦∈𝐶(4.3)
for all 𝑥∈𝐻. Then, the following holds: (1)𝑇𝑟 is single valued; (2)𝑇𝑟 is a firmly nonexpansive mapping, that is, for all 𝑥,𝑦∈𝐻,
‖‖𝑇𝑟𝑥−𝑇𝑟𝑦‖‖2≤⟨𝑇𝑟𝑥−𝑇𝑟𝑦,𝑥−𝑦⟩;(4.4)(3)𝐹(𝑇𝑟)=𝐸𝑃(𝐺); (4)𝐸𝑃(𝐺) is closed and convex.

We call such 𝑇𝑟 the resolvent of 𝐺 for 𝑟>0. Using Lemmas 4.1 and 4.2, we have the following lemma. See [34] for a more general result.

Lemma 4.3. Let 𝐻 be a Hilbert space and let 𝐶 be a nonempty, closed, and convex subset of 𝐻. Let 𝐺∶𝐶×𝐶→𝑅 satisfy (E1)–(E4). Let 𝐴𝐺 be a multivalued mapping of 𝐻 into itself defined by
𝐴𝐺𝑥={𝑧∈𝐻∶𝐺(𝑥,𝑦)≥⟨𝑦−𝑥,𝑧⟩,∀𝑦∈𝐶},∅,𝑥∈𝐶,𝑥∉𝐶.(4.5)
Then, 𝐸𝑃(𝐺)=𝐴𝐺−1(0) and 𝐴𝐺 is a maximal monotone operator with dom(𝐴𝐺)⊂𝐶. Further, for any 𝑥∈𝐻 and 𝑟>0, the resolvent 𝑇𝑟 of 𝐺 coincides with the resolvent of 𝐴𝐺; that is,
𝑇𝑟𝑥=(𝐼+𝑟𝐴𝐺)−1𝑥.(4.6)

Theorem 4.4. Let 𝐶 be a nonempty, closed, and convex subset of a real Hilbert space 𝐻. Let 𝐺 be a bifunction from 𝐶×𝐶→𝑅 satisfying (E1)–(E4) and let 𝑇𝜆 be the resolvent of 𝐺 for 𝜆>0. Suppose 𝐸𝑃(𝐺)≠∅. For 𝑢∈𝐶 and given 𝑥0∈𝐶, let {𝑥𝑛}⊂𝐶 be a sequence generated by
𝑥𝑛+1=𝛽𝑛𝑥𝑛+1−𝛽𝑛𝑇𝜆𝑛𝛼𝑛𝑢+1−𝛼𝑛𝑥𝑛(4.7)
for all 𝑛≥0, where {𝜆𝑛}⊂(0,∞), {𝛼𝑛}⊂(0,1), and {𝛽𝑛}⊂(0,1) satisfy (i)lim𝑛→∞𝛼𝑛=0 and ∑𝑛𝛼𝑛=∞;(ii)0<liminf𝑛→∞𝛽𝑛≤limsup𝑛→∞𝛽𝑛<1;(iii)𝑎≤𝜆𝑛≤𝑏 where [𝑎,𝑏]⊂(0,∞) and lim𝑛→∞(𝜆𝑛+1−𝜆𝑛)=0.Then {𝑥𝑛} converges strongly to a point ̃𝑥=𝑃𝐸𝑃(𝐺)(𝑢).